jmc

We have that $$\begin{array}{rl}{a}_2& =\sqrt{3}-2,\\ a_3& =(\sqrt{3}-2)\sqrt{3}-1=2-2\sqrt{3},\\ a_4& =(2-2\sqrt{3})\sqrt{3}-(\sqrt{3}-2)=\sqrt{3}-4,\\ a_5& =(\sqrt{3}-4)\sqrt{3}-(2-2\sqrt{3})=1-2\sqrt{3},\\ a_6& =(1-2\sqrt{3})\sqrt{3}-(\sqrt{3}-4)=-2,\\ a_7& =(-2)\sqrt{3}-(1-2\sqrt{3})=-1,\\ a_8& =(-1)\sqrt{3}-(-2)=2-\sqrt{3},\\ a_9& =(2-\sqrt{3})\sqrt{3}-(-1)=2\sqrt{3}-2,\\ a_{10}& =(2\sqrt{3}-2)\sqrt{3}-(2-\sqrt{3})=4-\sqrt{3},\\ a_{11}& =(4-\sqrt{3})\sqrt{3}-(2\sqrt{3}-2)=2\sqrt{3}-1,\\ a_{12}& =(2\sqrt{3}-1)\sqrt{3}-(4-\sqrt{3})=2,\\ a_{13}& =2\sqrt{3}-(2\sqrt{3}-1)=1.\end{array}$$Since $a_{12}=a_0=2$ and $a_{13}=a_1=1,$ and each term depends only on the previous two terms, the sequence is periodic from here on, with a period of length 12. Hence, $a_{100}=a_4=\boxed{\sqrt{3}-4}.$